MHT CET · Physics · Kinetic Theory of Gases
A monoatomic ideal gas is heated at constant pressure. The percentage of total heat used in changing the internal energy is
- A \(30 \%\)
- B \(40 \%\)
- C \(50 \%\)
- D \(60 \%\)
Answer & Solution
Correct Answer
(D) \(60 \%\)
Step-by-step Solution
Detailed explanation
For a monoatomic gas,
\(\gamma=\frac{C_{\mathrm{P}}}{\mathrm{C}_{\mathrm{V}}}=\frac{5}{3}\)
We know \(\Delta \mathrm{Q}=\mathrm{nC}_{\mathrm{p}} \Delta \mathrm{T}\) and \(\Delta \mathrm{U}=\mathrm{nC}_{\mathrm{v}} \Delta \mathrm{T}\)
\(\frac{\Delta U=\mathrm{nC}_{\mathrm{V}} \Delta \mathrm{~T}}{\Delta \mathrm{Q}=\mathrm{nC}_{\mathrm{p}} \Delta \mathrm{~T}}=\frac{\mathrm{C}_{\mathrm{V}}}{\mathrm{C}_{\mathrm{P}}}=\frac{3}{5}\)
\(\therefore \quad\) The percentage of total heat used in changing the internal energy is \(\frac{3}{5} \times 100=60 \%\).
\(\gamma=\frac{C_{\mathrm{P}}}{\mathrm{C}_{\mathrm{V}}}=\frac{5}{3}\)
We know \(\Delta \mathrm{Q}=\mathrm{nC}_{\mathrm{p}} \Delta \mathrm{T}\) and \(\Delta \mathrm{U}=\mathrm{nC}_{\mathrm{v}} \Delta \mathrm{T}\)
\(\frac{\Delta U=\mathrm{nC}_{\mathrm{V}} \Delta \mathrm{~T}}{\Delta \mathrm{Q}=\mathrm{nC}_{\mathrm{p}} \Delta \mathrm{~T}}=\frac{\mathrm{C}_{\mathrm{V}}}{\mathrm{C}_{\mathrm{P}}}=\frac{3}{5}\)
\(\therefore \quad\) The percentage of total heat used in changing the internal energy is \(\frac{3}{5} \times 100=60 \%\).
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